The lifespans of lizards in a particular zoo are normally distributed. The average lizard lives $3$ years; the standard deviation is $0.3$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a lizard living between $2.1$ and $2.4$ years.
Answer: $3$ $2.7$ $3.3$ $2.4$ $3.6$ $2.1$ $3.9$ $99.7\%$ $95\%$ $2.35\%$ $2.35\%$ We know the lifespans are normally distributed with an average lifespan of $3$ years. We know the standard deviation is $0.3$ years, so one standard deviation below the mean is $2.7$ years and one standard deviation above the mean is $3.3$ years. Two standard deviations below the mean is $2.4$ years and two standard deviations above the mean is $3.6$ years. Three standard deviations below the mean is $2.1$ years and three standard deviations above the mean is $3.9$ years. We are interested in the probability of a lizard living between $2.1$ and $2.4$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $99.7\%$ of the lizards will have lifespans within 3 standard deviations of the average lifespan. It also tells us that $95\%$ of the lizards will have lifespans within 2 standard deviations of the mean. That leaves $99.7\% - 95\% = 4.7\%$ of lizards between 2 and 3 standard deviations of the mean, or $2.35\%$ on either side of the distribution. The probability of a particular lizard living between $2.1$ and $2.4$ years is $\color{orange}{2.35\%}$.